Question

1g of a mixture of Al and Pb is placed in 50 ml of 1mol/L ZnCl2 solution. at the end of the reaction, 0.2 mol/L of Zn2+ ions were left
- write the equation(s) of the reaction(s)

Answer 1

(Pb + 2Al)_mix + 3ZnCl₂ (aq.)$\to$ 2AlCl₃(aq.) + (3Zn + Pb)_mix

Only Al reacts with ZnCl₂ because it is placed more reactive than Zn, while Pb is less reactive than Zn.

So, it does not participate in reaction.

According to data given, Initially 0.05 moles of Zn²⁺ ions are present.

Since, no. of moles = Molarity * Volume

After reaction, 0.01 moles are left. So, 0.04 moles are used up in reaction.

Therefore, according to stoichiometric ratio, 0.027 moles of Al are present in the mixture.

The amount of Al present in mixture = 0.72 g

Amount of substance = No. of moles * Molar Mass